Physics - Wave Optics Question with Solution | TestHub

Let intensity of two sources in YDSE be$I_1$and$I_2$. Assume phase difference between the sources is$\varphi .$

Resultant intensity in YDSE is given by

$I=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \varphi$.....(i)

It is given that,$a_1=2a_2$

$\Rightarrow a_1^2=4a_2^2$

$\Rightarrow I_1=4I_2$$\left[\because I\propto a^2\right]$.....(ii)

Now from Equations. (i) and (ii), we have

$I=4I_2+I_2+2\times 2\times I_2\cos \varphi$

$I=5I_2+4I_2\cos \varphi$.....(iii)

Given,$I_m=I_{\mathrm{m}\mathrm{a}\mathrm{x}}$

$={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2$

$\Rightarrow I_m={\left(2\sqrt{I_2}+\sqrt{I_2}\right)}^2$

$I_m=9I_2$

$I_2=\frac{I_m}{9}$......(iv)

Now from Equations. (iii) and (iv), we get

$I=5\times \left(\frac{I_m}{9}\right)+4\times \frac{I_m}{9}\mathrm{c}\mathrm{o}\mathrm{s}\varphi$

$=\frac{4I_m}{9}+\frac{4I_m}{9}\mathrm{c}\mathrm{o}\mathrm{s}\varphi +\frac{I_m}{9}$

$=\frac{4I_m}{9}\left[1+\cos \varphi \right]+\frac{I_m}{9}$

$=\frac{4I_m}{9}\left[2{\cos }^2\frac{\varphi }{2}\right]+\frac{I_m}{9}$

$\left[\because 1+\cos \varphi =2{\cos }^2\frac{\varphi }{2}\right]$

$=\frac{I_m}{9}\left[8{\cos }^2\frac{\varphi }{2}+1\right]$

$=\frac{I_m}{9}\left[1+8{\cos }^2\frac{\varphi }{2}\right]$